solve the given question (xlogx)dy/dx+y=logx^2
Answer (1)
18th Jan, 2021
Hello,
This question is very simple. Divide the equation by x. logx, on both sides.
So we get :
dy/dx + y/ x.logx = 2. logx / x .logx
i.e. dy/dx + y/ x logx = 2/x
Clearly this is a linear differential equation. So firstly we find the integrating factor.
I.F. = e ^ ( intg. { 1/ xlogx dx } ) ,
I.F. = logx
Now the solution of this differential equation is given by
y. I.F = intg. ( Q. I.F dx ) where Q = 2/x
Hence solution is : y. logx = (logx)^2 + c
Hope it helps.
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